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Exercise 7.6 · Q6

Q.Integrate the following function: x2log⁡xx^2 \log x

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The integral ∫x2log⁡x dx\int x^2 \log x \, dx is solved using integration by parts, treating log⁡x\log x as the first function and x2x^2 as the second. The result is x33log⁡x−x39+C\frac{x^3}{3} \log x - \frac{x^3}{9} + C.

The key here is recognising that log⁡x\log x and x2x^2 are not related by a simple substitution — one is logarithmic, the other a power. When you have a product of two different kinds of functions, integration by parts is the natural tool. The formula is:

∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du

The art lies in choosing which part is uu and which is dvdv. A reliable rule of thumb for Indian exams is the ILATE order: Inverse, Logarithmic, Algebraic, Trigonometric, Exponential. Whichever function comes first in ILATE should be taken as uu (the one you differentiate). Here, log⁡x\log x is Logarithmic, and x2x^2 is Algebraic. So log⁡x\log x comes first — it becomes uu, and x2x^2 becomes dvdv.

Let’s work through it.

  1. Set up the parts.

    Let u=log⁡xu = \log x and dv=x2 dxdv = x^2 \, dx.

    Differentiate uu: du=1x dxdu = \frac{1}{x} \, dx.

    Integrate dvdv: v=∫x2 dx=x33v = \int x^2 \, dx = \frac{x^3}{3}.

  2. Apply the integration by parts formula.

∫x2log⁡x dx=(log⁡x)⋅x33−∫x33⋅1x dx\int x^2 \log x \, dx = (\log x) \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx

Simplify the new integral: x33⋅1x=x23\frac{x^3}{3} \cdot \frac{1}{x} = \frac{x^2}{3}.

  1. Integrate the simpler term.

∫x23 dx=13∫x2 dx=13⋅x33=x39\int \frac{x^2}{3} \, dx = \frac{1}{3} \int x^2 \, dx = \frac{1}{3} \cdot \frac{x^3}{3} = \frac{x^3}{9}

  1. Combine everything. …

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